Chapter 29 : Balancing of Rotating Masses 587
29.2. METHODS FOR BALANCING OF ROTATING MASSES
We have already discussed that whenever a body of some mass is attached to a rotating shaft,
it exerts some centrifugal force, whose effect is to bend the shaft, and to produce unpleasant vibrations
in it. In order to prevent the effect of centrifugal force, another body is attached to the opposite side
of the shaft, at such a position, so as to balance the effect of centrifugal force of the first body. This
is done in such a way that the centrifugal force of both the bodies are made to be equal and opposite.
The process of providing the second body, in order to counteract the effect of the centrifugal force
of the first body, is called balancing of rotating masses.
29.3. TYPES OF BALANCING OF ROTATING MASSES
Though there are many types of balancing of rotating masses yet the following are important
from the subject point of view :
- Balancing of a single rotating mass.
- Balancing of several rotating masses.
29.4. BALANCING OF A SINGLE ROTATING MASS
A disturbing mass, attached to a rotating shaft, may be balanced in a number of ways. But the
following two are important from the subject point of view :
- Balancing by another mass in the same plane.
- Balancing by two masses in different planes.
29.5. BALANCING OF A SINGLE ROTATING MASS BY ANOTHER MASS IN
THE SAME PLANE
Fig. 29.1. Balancing of a single mass.
Consider a mass A, attached to a rotating shaft. In order to balance it, let us attach another mass
B to the same shaft as shown in Fig. 29.1.
Let m 1 = Mass of the body A,
r 1 = Radius of the rotation of body A (i.e., distance between the
centre of the shaft and the centre of the body A),
m 2 , r 2 = Corresponding values for the body B, and
ω = Angular velocity of the shaft.
We know that the centrifugal force exerted by the body A on the shaft
= m 1 ω^2 r 1 ...(i)
Similarly, centrifugal force exerted by the body B on the shaft
= m 2 ω^2 r 2 ...(ii)
Since the body B balances the body A, therefore the above two centrifugal forces should be
equal and opposite. Now equating (i) and (ii),
m 1 ω^2 r 1 = m 2 ω^2 r 2
∴ m 1 r 1 = m 2 r 2